AP EAMCET · Maths · Circle

The poles of the tangents to the circle $x^{2} + y^{2} = 4$ with respect to the circle $(x + 2)^{2} + y^{2} = 8$ lie on

A $y^{2} + 8 x = 0$
B $x^{2} + 8 y = 0$
C $y^{2} - 8 x = 0$
D $x^{2} - 8 y = 0$

Verified Solution

Answer & Solution

Correct Answer

(A) $y^{2} + 8 x = 0$

Step-by-step Solution

Detailed explanation

Given: x2+y2=4 x+22+y2=8 Equation of tangent at any point on the circle x2+y2=4 is xcosθ+ysinθ=r xcosθ+ysinθ=2 ...1 The polar of x1,y1 w.r.t. x+22+y2=8 xx1+yy1+2x+x1-4=0 ⇒ x1+2x+y1y+2x1-2=0 ...2 from equation…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From AP EAMCET

List-I (Compound)	List-II (Use)
A)	Kieselghur	I)	Chromatographic material
B)	Silica gel	II)	Softening of hard water
C)	ZSM-5	III)	Filtration plants
D)	Hydrated zeolites	IV)	To convert alcohol directly into gasoline

Explore more questions on app

The poles of the tangents to the circle x2+y2=4 with respect to the circle (x+2)2+y2=8 lie on

Step-by-step Solution

The poles of the tangents to the circle $x^{2} + y^{2} = 4$ with respect to the circle $(x + 2)^{2} + y^{2} = 8$ lie on